The Bates model is represented by the bivariate system of stochastic differential equations
\begin{align}
&dS_t=(r-q)S_tdt+\sqrt{v_t}S_tdW_1(t)+S_tdN_t\\
&dv_t=\kappa(\theta-v_t)dt+\sigma\sqrt{v_t}dW_2(t)\\
\end{align}
where
$$\mathbb{E^Q}[dW_1(t)dW_2(t)]=\rho dt$$
and $N_t$ is a compound Poisson process with intensity $\lambda$ and independent jumps $J$ with
$$\ln{(1+J)}\tilde{~} N\left(\ln(1+\beta)-\frac{1}{2}\alpha^2\,\, ,\,\,\alpha^2\right)$$
The parameters $\beta$ and ${\alpha}$ determine the distribution of the jumps and the Poisson process is assumed to be independent of the Wiener processes.By application of delta-hedging argument,we have
$$\frac{\partial U}{\partial t}+\frac{1}{2}{{v}_{t}}{{S}_{t}}^{2}\frac{{{\partial }^{2}}U}{\partial {{S}^{2}}}+\frac{1}{2}{{\sigma }^{2}}{{v}_{t}}\frac{{{\partial }^{2}}U}{\partial {{v}^{2}}}+\rho \sigma \,{{v}_{t}}{{S}_{t}}\frac{{{\partial }^{2}}U}{\partial S \partial v}+\,(r-q-\lambda\,\bar{k}){{S}_{t}}\frac{\partial U}{\partial S}+\kappa (\theta -{{v}_{t}})\,\frac{\partial U}{\partial v}-rU+I_U=0\,\,\,(1)$$
where
$$I_U=\lambda\int_{0}^{\infty}[U(S\xi,v,t)-U(S,v,t)]\,g(\xi)\,d\xi$$
$$g(\xi)=\frac{1}{\sqrt{2\pi}\alpha\xi}e^{-\frac{1}{2\alpha^2}(\ln{\xi-m})^2}$$
$$m=\ln(1+\beta)-\frac{1}{2\alpha^2}$$
$$\bar{k}=e^{\frac{1}{2}\alpha^2+m}-1$$
It should be noted that closed-form solutions for vanilla-option payoff do exist but PIDE (1) is easily approximated by Numerical Methods.
close form solution
I edited my Answer for emcor.
- Dynamic of $S_t$ under historical measure
\begin{align}
&\frac{dS_t}{S_t}=(\mu-\lambda\,\bar{J})dt+\sqrt{v_t}dW_1(t)+J\,dN(t)\\
&\hspace{0.3cm}dv_t=\kappa(\theta-v_t)dt+\sigma\sqrt{v_t}dW_2(t),\\
\end{align}
where
$$\mathbb{E^P}[dW_1(t)dW_2(t)]=\rho dt,$$
$N_t$ is a compound Poisson process with intensity $\lambda$ and independent jumps $J$ with
$$\ln{(1+J)}\tilde{~} N\left(\ln(1+\bar{J})-\frac{1}{2}\alpha^2\,\, ,\,\,\alpha^2\right).$$
The parameters $\bar{J}$ and ${\alpha}$ determine the distribution of the jumps and the Poisson process is assumed to be independent of the Wiener processes.
- Change measure: $\mathbb{P\rightarrow Q}$
\begin{align}
&\frac{dS_t}{S_t}=(r-q-\lambda^*\,\bar{J^*})dt+\sqrt{v_t}dW_1^{\mathbb{Q}}(t)+J^*\,dN^*(t)\\
&\hspace{0.3cm}dv_t=\kappa^*(\theta^*-v_t)dt+\sigma\sqrt{v_t}dW_2^{\mathbb{Q}}(t),\\
\end{align}
where
$$\mathbb{E^Q}[dW_1^{\mathbb{Q}}(t)dW_2^{\mathbb{Q}}(t)]=\rho dt$$
$$\kappa^*=\kappa +\xi$$
$$\hspace{0.3cm}\theta^*=\frac{\kappa\theta}{\kappa+\xi},$$
such that $\xi$ is volatility market price and
$$J^*=J+J\,\mathbb{E^P}\left[\frac{\Delta J_w}{J_w}\right]$$
$$\bar{J}^*=\bar{J}+\frac{\mathbb{Cov}\left(J,\frac{\Delta J_w}{J_w}\right)}{1+\mathbb{E^P}\left[\frac{\Delta J_w}{J_w}\right]}.$$
Where $\frac{\Delta J_w}{J_w}$ is random percentage jump conditional on a jump occurring and $\frac{dJ_w}{J_w}$ is percentage shock in the absence of jump.
- Note that,when $\xi=0$ we have $\kappa^*=\kappa$ and $\theta^*=\theta$. We set $\xi=0$, because when we estimate the risk-neutral parameters to price options we do not need to estimate $\xi$. Also, when $\Delta J_w/J_w\rightarrow0$ thus we have $J^*=J$ and $\bar{J}^*=\bar{J}.$
- let $x_t=\ln S_t$ then
\begin{align}
C(t\,,{{S}_{t}},{{v}_{t}},J,K,T)={{S}_{t}}{{P}_{1}}-K\,{{e}^{-r\tau }}{{P}_{2}}
\end{align}
where,for $j=1,2$
\begin{align}
& {{P}_{j}}({{x}_{t}}\,,\,{{v}_{t}}\,;\,\,{{x}_{T}},\ln K)=\frac{1}{2}+\frac{1}{\pi }\int\limits_{0}^{\infty }{\operatorname{Re}\left( \frac{{{e}^{-i\phi \ln K}}{{f}_{j}}(\phi ;t,x,v)}{i\phi } \right)}\,d\phi \\
&\\
&\hspace{1.9cm}{{f}_{j}}(\phi ;{{v}_{t}},{{x}_{t}})=\exp\left[{{C}_{j}}(\tau ,\phi) +{{D}_{j}}(\tau ,\phi ){{v}_{t}}+i\phi{{x}_{t}}+{{\Xi }_{j}}\right]\\
&\\
&\hspace{3.8cm}\Xi_j={{\lambda }^{*}}\tau\,{{(1+{{\kappa }^{*}})}^{{{u}_{j}}+\frac{1}{2}}}\left[ {{(1+{{\kappa }^{*}})}^{\phi }}{{e}^{{{\alpha }^{2}}({{u}_{j}}\phi +0.5{{\phi }^{2}})}}-1 \right],\\
\end{align}
such that
\begin{align}
&C_j(\tau ,\phi)=(r-q-\lambda^*\bar{\kappa}^*)\phi\,\tau-\frac{\kappa^*\theta^*\,\tau}{\sigma^2}(\rho\sigma\phi-\beta_j-\gamma_j)-\frac{2\kappa^*\theta^*\,\tau}{\sigma^2}\ln\left(1+\frac{1}{2}(\rho\sigma\phi-\beta_j-\gamma_j)\frac{1-e^{\gamma_j\tau}}{\gamma_j}\right)\\
&\\
&D_j(\tau ,\phi)=\frac{-2(u_j\phi+\frac{1}{2}\phi^2)}{\rho\sigma\phi-\beta_j+\gamma_j\frac{1+e^{1-\gamma_j\tau}}{1-e^{1-\gamma_j\tau}}}\\
&\\
&\hspace{1.1cm}\gamma_j=\sqrt{(\rho\sigma\phi-\beta_j)^2-2\sigma^2(u_j\phi+0.5\phi^2)}\\
\end{align}
and
$$u_1=\frac{1}{2}\,,u_2=-\frac{1}{2}\,,\beta_1=\kappa^*-\rho\sigma\,,\beta_2=\kappa^*\,$$
